Advertisements
Advertisements
प्रश्न
Evaluate the following limits:
`lim_(x -> 0) (sin^3(x/2))/x^2`
Advertisements
उत्तर
We know `lim_(x -> 0) sinx/x` = 1
`lim_(x -> 0) (sin^3(x/2))/x^3 = lim_(x -> 0) (sin^3(x/2))/(2^3 xx x^3/2^3)`
= `lim_(x/2 -> 0) 1/8* (sin^3(x/2))/(x/2)^3`
= `1/8 lim_(x/2 -> 0) [(sin (x/2))/(x/2)]^3`
`lim_(x -> 0) (sin^3(x/2))/x^2 = 1/8 xx 1^3`
= `1/8`
APPEARS IN
संबंधित प्रश्न
Evaluate the following limit:
`lim_(z -> -5)[((1/z + 1/5))/(z + 5)]`
Evaluate the following limit:
`lim_(x -> 5)[(x^3 - 125)/(x^5 - 3125)]`
Evaluate the following limit :
`lim_(x -> 0)[((1 - x)^8 - 1)/((1 - x)^2 - 1)]`
Evaluate the following :
`lim_(x -> 0)[x/(|x| + x^2)]`
Evaluate the following :
Find the limit of the function, if it exists, at x = 1
f(x) = `{(7 - 4x, "for", x < 1),(x^2 + 2, "for", x ≥ 1):}`
Evaluate the following :
`lim_(x -> 0) {1/x^12 [1 - cos(x^2/2) - cos(x^4/4) + cos(x^2/2) cos(x^4/4)]}`
Verify the existence of `lim_(x -> 1) f(x)`, where `f(x) = {{:((|x - 1|)/(x - 1)",", "for" x ≠ 1),(0",", "for" x = 1):}`
Evaluate the following limits:
`lim_(x -> 2) (1/x - 1/2)/(x - 2)`
Evaluate the following limits:
`lim_(x -> 2) (2 - sqrt(x + 2))/(root(3)(2) - root(3)(4 - x))`
Evaluate the following limits:
`lim_(x -> 0) (sqrt(1 - x) - 1)/x^2`
Evaluate the following limits:
`lim_(x ->oo) (x^3/(2x^2 - 1) - x^2/(2x + 1))`
Evaluate the following limits:
`lim_(x -> oo)(1 + "k"/x)^("m"/x)`
Evaluate the following limits:
`lim_(x -> 0) (tan 2x)/(sin 5x)`
Evaluate the following limits:
`lim_(x -> 0) (1 - cos^2x)/(x sin2x)`
Evaluate the following limits:
`lim_(x -> 0) ("e"^x - "e"^(-x))/sinx`
Evaluate the following limits:
`lim_(x -> 0) ("e"^("a"x) - "e"^("b"x))/x`
Choose the correct alternative:
`lim_(x -> oo) sinx/x`
Choose the correct alternative:
`lim_(alpha - pi/4) (sin alpha - cos alpha)/(alpha - pi/4)` is
`lim_(x -> 0) ((2 + x)^5 - 2)/((2 + x)^3 - 2)` = ______.
`lim_(x -> 0) (sin 4x + sin 2x)/(sin5x - sin3x)` = ______.
